# How do you integrate int sqrt((x+3)^2-100) using trig substitutions?

Nov 2, 2016

The answer is $= \frac{x + 3}{2} \left(\sqrt{\left({\left(x + 3\right)}^{2} - 100\right)}\right) - 50 \ln \left(\frac{\sqrt{\left({\left(x + 3\right)}^{2}\right) - 100} + \left(x + 3\right)}{10}\right) + C$

#### Explanation:

Let $u = x + 3$ then $\mathrm{du} = \mathrm{dx}$
$\int \left(\sqrt{{\left(x + 3\right)}^{2} - 100}\right) \mathrm{dx} = \int \left(\sqrt{{u}^{2} - 100}\right) \mathrm{du}$
Then let $u = 10 \sec \theta$$\implies$$\mathrm{du} = 10 \sec \theta \tan \theta$
$\int \left(\sqrt{{u}^{2} - 100}\right) \mathrm{du} = \int \left(\sqrt{100 {\sec}^{2} \theta - 100}\right) 10 \sec \theta \tan \theta \left(d \left(\theta\right)\right)$
$= 100 \int \sec \theta {\tan}^{2} \theta d \left(\theta\right)$
$= 100 \int \sec \theta \left({\sec}^{2} \theta - 1\right) d \left(\theta\right)$
$= 100 \int \left({\sec}^{3} \theta - \sec \theta\right) d \left(\theta\right)$
$\int {\sec}^{3} \theta d \left(\theta\right) = \frac{1}{2} \int \sec \theta d \left(\theta\right) + \frac{1}{2} \sec \theta \tan \theta$
and $\int \sec \theta d \left(\theta\right) = \ln \left(\tan \theta + \sec \theta\right)$
$\int {\sec}^{3} \theta d \left(\theta\right) = \frac{1}{2} \ln \left(\tan \theta + \sec \theta\right) + \frac{1}{2} \sec \theta \tan \theta$
$100 \int \left({\sec}^{3} \theta - \sec \theta\right) d \left(\theta\right) = 100 \left(\frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln \left(\tan \theta + \sec \theta\right)\right)$